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Problem :
Find the critical points and inflection points of the function f (x) = x^{4} -2x^{2} (with domain
the set of all real numbers). Which of the critical points are local minima? local
maxima? Is there a global minimum or maximum?

We first calculate the derivatives of the function:

f'(x)

=

4x^{3} - 4x

=

4(x + 1)x(x - 1)

f''(x)

=

12x^{2} - 4

=

4(3x^{2} - 1)

We see that f'(x) = 0 when x = - 1, 0, or 1, so these are the three critical points of
f. We calculate the second derivatives at these points:

f''(- 1)

=

8

f''(0)

=

-4

f''(1)

=

8

so by the second derivative test, f has local minima at -1 and 1 and a local maximum
at 0. Substituting back into the original function yields

f (- 1)

=

-1

f (0)

=

0

f (1)

=

-1

so f attains its global minimum of -1 at x = ±1. It is clear from the graph of f
that it has no global maximum.
To find the points of inflection, we solve f''(x) = 0, or 12x^{2} - 4 = 0, which has
solutions x = ±1/3) ±0.58. Referring once again to the graph of f,
we can check that the concavity does indeed change at these x-values.

Problem :
Find the inflection points of f (x) = e^{-x2}. (This famous function
is called the gaussian.)

We compute the derivatives:

f'(x)

=

-2xe^{-x2},

f''(x)

=

(- 2x)(- 2xe^{-x2}) + (- 2)(e^{-x2})

=

(4x^{2} -2)e^{-x2}

Solving f''(x) = 0 for x, we get the inflection points x = ±1/ 0.71.
This is reasonable when one considers the graph of f.

Problem :
Find a function f (x) with inflection points at x = 1 and x = 2.

We need f''(1) = f''(2) = 0, so we might as well let

f''(x) = (x - 1)(x - 2) = x^{2} - 3x + 2

We search for a function with this derivative, ultimately obtaining

f'(x) = - + 2x

Similarly, a function with this derivative is given by